-
1) i) f(t) = e2t sin(t).The transform of sin(t) is
1s2 + 1
,
and when we multiply by an exponential we need to shift where
this transformis evaluated. Thus,
F (s) =1
(s 2)2 + 1 .
ii) f(t) = u(t 2)(t2 2t).To transform this, we will need to
write the coefficient function t2 2t in
the form h(t 2), for some function h(t). Once we have this, F
(s) = e2sH(s),where H(s) is the Laplace transform of h(t). So,
t2 2t = (t 2 + 2)2 2(t 2 + 2)= (t 2)2 + 4(t 2) + 4 2(t 2) 4= (t
2)2 + 2(t 2)= h(t 2).
Thus h(t) = t2 + 2t, and
F (s) = e2s(
2t3
+2t2
).
iii) f(t) = 2t3et2
This is similar to i); the Laplace transform of 2t3 is
12s4,
which we have to shift by 12 . Thus,
F (s) =12(
s+ 12)4 .
iv) f(t) = u(t pi2 )et sin(t).Just like in ii), we need to force
the shift in the coefficient function.
et sin(t) = etpi2+
pi2 sin
(t pi
2+pi
2
)= e
pi2 et
pi2 cos
(t pi
2
)= h
(t pi
2
)Thus h(t) = e
pi2 et cos(t), and
F (s) = epi2 se
pi2
s 1(s 1)2 + 1 .
1
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2) i) F (s) = e2s 2s3We begin by inverse transforming 2s3 ,
which has an inverse transform of t
2.Then, to get f(t), we need to shift this by 2 and multiply it
by u(t 2), whichgives us an inverse transform of
f(t) = u(t 2)(t 2)2.
ii) F (s) = 4s26s+10The first thing to notice is that the
denominator doesnt factor; rather, we
complete the square to get s2 6s+ 10 = (s 3)2 + 1. So the
inverse transformf(t) will be
f(t) = 4e3t sin(t).
iii) F (s) = 3s+5(s+1)2+16Looking at the denominator, we see
that the terms in the inverse transform
will involve et sin(4t) and et cos(4t). We just need to fix up
the numeratorsto match. We start by forcing the s to become (s +
1), and then get the rightnumerator of 4 for the constant term.
F (s) =3(s+ 1 1) + 5
(s+ 1)2 + 16
=3(s+ 1) + 2(s+ 1)2 + 16
=3(s+ 1)
(s+ 1)2 + 16+
12
4(s+ 1)2 + 16
Sof(t) = 3et cos(4t) +
12et sin(4t).
iv) F (s) = 5s216s+6
s(s2)(s3)In this case, the denominator is factored, so well need
to do partial fractions.
Doing so gives
F (s) =1s
+35
1s 2
15
1s 3 .
Thus the inverse transform is
f(t) = 1 +35e2t 1
5e3t.
3) i) y + 8y + 52y = 13u(t 2), y(0) = 0, y(0) = 1Transforming
the entire equation and plugging in the initial conditions
gives
(s2 + 8s+ 52)Y (s) 1 = 13se2s
Y (s) =1
s2 + 8s+ 52+ e2s
13s(s2 + 8s+ 52)
.
2
-
The first term can be dealt with directly, as s2 + 8s + 52 = (s
+ 4)2 + 36, butfirst, we should partial fraction the second term.
After doing so, we get
Y (s) =1
(s+ 4)2 + 36+ e2s
(14
1s 1
4s
(s+ 4)2 + 36 2
(s+ 4)2 + 36
)=
16
6(s+ 4)2 + 36
+ e2s(
14
1s 1
4s+ 4 4
(s+ 4)2 + 36 2
(s+ 4)2 + 36
)=
16
6(s+ 4)2 + 36
+ e2s(
14
1s 1
4s+ 4
(s+ 4)2 + 36 1
66
(s+ 4)2 + 36
),
and so the solution is
y(t) =16e4t sin(6t)+u(t2)
(14 1
4e4(t2) cos(6(t 2)) 1
6e4(t2) sin(6(t 2))
).
ii) y + 9y = 3 3u(t 1), y(0) = 1, y(0) = 0.Transforming the
whole equation and solving for Y (s) gives us
(s2 + 9)Y (s) + s =3s es 3
s
Y (s) = ss2 + 9
+3
s(s2 + 9 es 3
s(s2 9)= s
s2 + 9+
13
1s 1
3s
s2 + 9 es
(13
1s 1
3s
s2 + 9
),
so the solution is
y(t) = 43
cos(3t) +13 u(t 1)
(13 1
3cos(3(t 1))
).
iii) y + 4y = 1 2t, y(0) = 1, y(0) = 2Transforming the whole
equation and solving for Y (s) gives
Y (s) =s+ 6s(s+ 4)
+1
s2(s+ 4) 2s3(s+ 4)
=s3 + 6s2 + s 2
s3(s+ 4),
which, after doing partial fractions, breaks up into
=4532
1s
+38
1s2 1
21s3 13
321
s+ 4,
so that the solution is
y(t) =4532
+38t 1
4t2 13
32e4t.
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